Question on Spacetime dilation.
in [Relativity]
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From: Ken S. Tucker on 22 Dec 2008 04:09 Hi John. On Dec 21, 8:09 pm, John Polasek <jpola...(a)cfl.rr.com> wrote: > On Thu, 18 Dec 2008 14:12:06 -0800 (PST), "Ken S. Tucker" > > > > <dynam...(a)vianet.on.ca> wrote: > >Hi XXein and guys.... > >... > >> xxein: Welcome aboard to both you and Allen. I hope you enjoy the > >> ride. > >I posted this to SPF... > > >To Theo and all theoreticians who embrace the signature (+---), > >I'm unable to understand it so far, but I'm trying! > > >On Dec 18, 12:16 pm, Theo Wollenleben <alpha0...(a)yahoo.de> wrote: > >> Ken S. Tucker schrieb: > >> > On Dec 16, 2:50 pm, Theo Wollenleben <alpha0...(a)yahoo.de> wrote: > >> >> Ken S. Tucker schrieb: > >> >>> ds^2 =g_uv dx^u dx^v , {u,v=0,1,2,3}, > >> >>> = dx_u dx^u , > >> >>> = dx_0 dx^0 + dx_i dx^i , {i=1,2,3} , Eq.(1). > >> >>> dt' = ds = (0.6) dt, Eq.(2). > >> >>> What differential coefficients should be subbed > >> >>> into Eq.(1) to yield Eq.(2)? > > >> >> g_uv = diag(1,-1,-1,-1) > >> >> dx_0 = dx^0, dx_i = -dx^i > >> >> ds^2 = (dx^0)^2 - (dx^i)^2 > >> >> ds = sqrt(1 - (dx^i/dx^0)^2)*dx^0 = sqrt(1-(v/c)^2)*cdt > > >> > If "dr" is a spatial displacement *vector*, and > >> > e^i and e_i are the 3D spatial basis vectors, the > >> > text book suggests that, > >> > dr = e_i dx^i = e^i dx_i . > >> > and also > >> > dr^2 = dr.dr , (that's a dot/scalar product) > >> > = e_i.e_j dx^i dx^j = g_ij dx^i dx^j > >> > = e^i.e^j dx_i dx_j = g^ij dx_i dx_j > >> > does that seem reasonable to you Theo and all? > > >> Where the scalar product is given by x.y=g(x,y). That is correct, though > >> "being spatial" in Minkowski space is not an invariant concept, of > >> course. If we choose another basis, the same vector could have a nonzero > >> time component. > > >Ok, I've studied metrics such as g_i0, however they are not popular, > >and I have been advised to understand the signature (1,-1,-1,-1) > >you (Theo) introduced above, that are conventionally used by most > >relativists and what they mean by that. > > >A minor problem I'm stuck on is this, if > > >dr = e_i dx^i = e^i dx_i , and dx^i = - dx_i then e_i = -e^i . > > >The tensor textbook requires that the Kronecker Delta " delta^u_v " > >has values of 1,0 such as > > >delta^u_v = {1,0}when {u=v , u=/=v}, defined by > > >delta^u_v = e^u.e_v , but > > I doubt it. Well I have no doubt, one of us knows what I'm doing. > The LHS is second rank, and the RHS 0 rank. The equation > has no value. Well that's easy, dismiss the foundation of tensor analysis because you don't understand it! What YOU need to prove, when adopting the signature (+---) is that it will NOT generate 3 values for the Kronecker delta, ( +1 0 -1), and if you cannot do that, prove a 3 valued Kronecker in any CS is logically valid, and if you cannot do that, then that specialized CS is not valid. What do you think mathematical physics is, a circus ?? Back in the 60's and 70's, in consultation with a few of the best GR theoreticians this planet has seen, privately informed me that the (+---) was awkward at best and likely wrong. And furthermore, tasked me to straighten out the "mess". At that time NO ONE could maintain a consistency from basis vectors (unitary base vectors) directly to the GR metric boundary. In the absence of that solution, that I've requested in this group and also, SPF and spr, that no one knows now (2008), and back then 1970's, we decided by 1983 to solve the "mess" by interdefining the meter and second, for a brief see, http://physics.trak4.com/modern-spacetime.pdf and a mathematically and physically proven signature becomes (++++) in accord with the ISU, following my tasked results. I did the task in the late 70's, published the results, the ISU accepted them and then redefined the meter, in 1983, and no one paid much attention to that, which at the time was reasonably quite unimportant to most people. It would have remained unimportant until a test of the Kerr metric was conceived and brought to fruitation by satellite experiment, the GP-b, (something NASA argued against, but were politically pressured to support). So now Tucker is back in main-stream, (I'm ok with that), explaining the impact of the 1983 decision on the Kerr metric and the resulting null "frame dragging" here, http://physics.trak4.com/MST_Kerr.pdf It appears to me in hind-sight, by adopting signature (+---) will suggest a Kerr metric with antisymmetric rotational artifacts from the "g_01" metric fed-back into ds^2, for ref see Weinberg's "Grav&Cosmo" pg 240, that was developed from the (+---) signature. OTOH, dating back to 1983, my ref above excludes the Kerr metric using the (++++), signature, funny how things turn out... The GP-b becomes an experiment to determine which signature is correct, like a battle between (++++) and (+---), waged in outer space by two opposing groups of scientists, by a satellite with Tucker at the fulcrum, but I digress... > >e^1.e_1 = -1 , e^0.e_0 = +1 , e^1.e_0 = 0 , etc. Yes John, I look forward to your improvements. Regards Ken S. Tucker
From: Alen on 22 Dec 2008 08:57 On Dec 22, 2:38 am, "Ken S. Tucker" <dynam...(a)vianet.on.ca> wrote: > Hi Alen. > > On Dec 20, 6:14 pm, Alen <al...(a)westserv.net.au> wrote: > > > On Dec 20, 12:21 pm, "Ken S. Tucker" <dynam...(a)vianet.on.ca> wrote: > > [...] > > > Maybe Alen is a math genious, and knows more > > > than Tom does, but that's easy to do anyway. > > > Regards > > > Ken S. Tucker > > > I don't think I have to be a genius, and don't see > > what the fuss is about, although I know that Tom > > always likes to insist on 'absolute' precision in > > notation. Even so, suppose we represent a vector > > x + iy as A^i and define > > > A^iA_i = 1, > > then we will have > > A_i = (1/|A^i|^2)(x - iy) > > > If you multiply them you will get 1. Therefore we > > should be able to meaningfully say > > A_i = 1/A^i > > > since we can say > > A_iX = X/A^i and, if X = A^i, > > A_iA^i = A^i/A^i = 1 > > Careful, even vanilla Minkowski spacetime has > sqrt(-1) all over the damn place. > > Set A_i = i * |A_i| and A^i = i * |A^i|, > > (Does the Einstein summation convention apply?). I wasn't doing any summation here. I was merely indicating, by an example, that the concept of an 'inverse' can be made to work with realities more complicated than just simple numbers. > > > There is therefore some kind of meaningful > > 'inverse' relationship between A^i and A_i, > > in this case, and to realise this is helpful to > > complete an understanding of the nature of > > their relationship > > > Similarly, if we have > > x^ix_j = d^i_j, with d for delta, > > > we will have > > x^ix_i = 1 > > Is the summation convention suspeneded there? It is not being discussed. The above is discussing only individual values of i. That is, if we say d^i_j is 0, if i != j, and, 1 if i = j, we are referring only to individual pairs of values of i and j. > > and this therefore 'must' always confer 'some' > > kind of valid meaning on > > x_i = 1/x^i > > > 'whatever' kind of an object x might refer to. 'Some' > > kind of inverse is always a conceptual counterpart > > of the simple fact that x^ix_i = 1. Is that not so?? > > The notation is a bit vague, I'd be happy to see > time + 1D such as 0,1 in spacetime. > > > Alen > > Regards > Ken S. Tucker The generalised concept of an inverse is, perhaps, vague but, then, even a numerical inverse is less clear than a simple number. 6, for example, can directly refer to 6 existing objects. So can 3. 1/2 of 6, or 6/2, however, doesn't directly refer to anything in a simple way, but indicates, rather, a process or, more accurately, a more complicated process than a simple count. This process involves a comparison between two distinct simple realities, i.e. 6 objects unified, and two sets of 3 objects. So an inverse is not a simple concept in any case, and has more of the quality of an operator than a simple number. That is what I think. I don't think I can help with Minkowski spacetime, however! Alen
From: John Polasek on 22 Dec 2008 11:51 On Sun, 21 Dec 2008 22:05:16 -0800 (PST), "Ken S. Tucker" <dynamics(a)vianet.on.ca> wrote: >On Dec 21, 8:00 pm, John Polasek <jpola...(a)cfl.rr.com> wrote: >> On Tue, 16 Dec 2008 11:57:32 -0800 (PST), "Ken S. Tucker" >> >> <dynam...(a)vianet.on.ca> wrote: >> >A clock (in K) moving at 0.8c (relative to K') is >> >dilated 0.6 by t' = t*sqrt(1 - v^2/c^2), so that >> >t'=(0.6)*t. >> >> >In GR that is generalized to be, >> >> >ds^2 =g_uv dx^u dx^v , {u,v=0,1,2,3}, >> >> >and then by association equatable to >> >> >= dx_u dx^u , >> >> >= dx_0 dx^0 + dx_i dx^i , {i=1,2,3} , Eq.(1). >> >> Firstly, you are incorporating .8c which is related to SR, and >> admixing it into GR. > >"admixing", SR is a GR limit. > >> Secondly, if you retain GR's guv matrix, even with zero gravity, then >> your first term should be negative, that is, -dx0dx0. > >WHY? We want logic, NOT idle pronouncements. > >> I think I'm right on this. > >Prove it. g_00 = -1 + 2GM/rc^2 = -1 +eps, in the limit, -1 g_11 = 1/(1 -eps), in the limit, 1. dx_0 g00 dx^0 = -dx0^2 in the limit >>Remember the signature -1 1 1 1. > >That's a conjecture I want to see proven. >Regards >Ken S. Tucker John Polasek
From: Ken S. Tucker on 22 Dec 2008 13:25 HI Alen. On Dec 22, 5:57 am, Alen <al...(a)westserv.net.au> wrote: > On Dec 22, 2:38 am, "Ken S. Tucker" <dynam...(a)vianet.on.ca> wrote: > > On Dec 20, 6:14 pm, Alen <al...(a)westserv.net.au> wrote: > > > > On Dec 20, 12:21 pm, "Ken S. Tucker" <dynam...(a)vianet.on.ca> wrote: > > > [...] > > > > Maybe Alen is a math genious, and knows more > > > > than Tom does, but that's easy to do anyway. > > > > Regards > > > > Ken S. Tucker > > > > I don't think I have to be a genius, and don't see > > > what the fuss is about, although I know that Tom > > > always likes to insist on 'absolute' precision in > > > notation. Even so, suppose we represent a vector > > > x + iy as A^i and define > > > > A^iA_i = 1, > > > then we will have > > > A_i = (1/|A^i|^2)(x - iy) > > > > If you multiply them you will get 1. Therefore we > > > should be able to meaningfully say > > > A_i = 1/A^i > > > > since we can say > > > A_iX = X/A^i and, if X = A^i, > > > A_iA^i = A^i/A^i = 1 > > > Careful, even vanilla Minkowski spacetime has > > sqrt(-1) all over the damn place. > > > Set A_i = i * |A_i| and A^i = i * |A^i|, > > > (Does the Einstein summation convention apply?). > > I wasn't doing any summation here. I was merely > indicating, by an example, that the concept of an > 'inverse' can be made to work with realities more > complicated than just simple numbers. > > > There is therefore some kind of meaningful > > > 'inverse' relationship between A^i and A_i, > > > in this case, and to realise this is helpful to > > > complete an understanding of the nature of > > > their relationship > > > > Similarly, if we have > > > x^ix_j = d^i_j, with d for delta, > > > we will have > > > x^ix_i = 1 > > Is the summation convention suspended there? > It is not being discussed. The above is discussing > only individual values of i. That is, if we say d^i_j is 0, > if i != j, and, 1 if i = j, we are referring only to individual > pairs of values of i and j. > > > > and this therefore 'must' always confer 'some' > > > kind of valid meaning on > > > x_i = 1/x^i > > > > 'whatever' kind of an object x might refer to. 'Some' > > > kind of inverse is always a conceptual counterpart > > > of the simple fact that x^ix_i = 1. Is that not so?? > > > The notation is a bit vague, I'd be happy to see > > time + 1D such as 0,1 in spacetime. > > > > Alen > > > Regards > > Ken S. Tucker > > The generalised concept of an inverse is, perhaps, vague > but, then, even a numerical inverse is less clear than > a simple number. 6, for example, can directly refer to 6 > existing objects. So can 3. 1/2 of 6, or 6/2, however, doesn't > directly refer to anything in a simple way, but indicates, > rather, a process or, more accurately, a more complicated > process than a simple count. This process involves a comparison > between two distinct simple realities, i.e. 6 objects unified, > and two sets of 3 objects. So an inverse is not a simple > concept in any case, and has more of the quality of an > operator than a simple number. That is what I think. Well I think you are right. I rechecked the definition of a unit vector in curvilinear coordinates, I'll copy it for you, e1 is the unit vector, r is a position vector, (I use & for partial, "|" are mag bars), e1 = &r/&x1 / |&r/&x1| It seems entirely sensible to me to set 1 / e1 = |&r/&x1| / &r/&x1 because the RHS is defined and then to the LHS. > I don't think I can help with Minkowski spacetime, however! Well for now I'll stay with the tried and true Modern SpaceTime, until Minkowski spacetime is proven from the ground up to be consistent, by our standards. > Alen Regards Ken S. Tucker
From: Ken S. Tucker on 22 Dec 2008 16:10
Hey I figured out the space part of the signature (+---), what do you think ? e_1 = i , e^1 = -i , e_1 = - e^1 , ok. g_11 = e_1 . e_1 = (i) (i) = -1 , ok g^1_1 = e^1.e_1 = (-i) (i) = +1 , ok g^11 = e^1.e^1 = (-i) (-i) = -1 , ok Regards Ken S. Tucker |